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On page 2 of Dummit & Foote's Abstract Algebra (3rd ed.), Proposition 1/(4) states:

(Let $f:A \to B$.) If $A$ and $B$ are finite sets with the same number of elements (i.e., $\lvert A \rvert = \lvert B \rvert$), then $f:A \to B$ is bijective if and only if $f$ is injective if and only if $f$ is surjective.

Firstly, may I confirm that empty set is treated as finite set?

If so, let $A = B = \emptyset$, then we have $\lvert A \rvert = \lvert B \rvert = 0$.

Secondly, may I ask if we could talk about function from empty set to empty set?

If so, to me the definition of such function goes like "for each $x\in A$, there exists a unique $b\in B$ being assigned with x". And this should be vacuously true. So the function is well-defined.

Now, the statement "for each $a,b\in A$, if $a\neq b$, then $f(a)\neq f(b)$" is vacuously true, so the function is injective. Meanwhile, the statement "for each $b\in B$, there exists $a\in A$ such that $f(a)=b$" is vacuously true, so the function is surjective and thus bijective.

Finally, may I ask if this completes the proof for the $\emptyset$ case of Proposition 1/(4)?

Deltaspace
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J-A-S
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1 Answers1

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Yes, the empty function, from the empty set to the empty set, is vacuously bijective, and your proof is fine.

In fact, any function with the empty set as domain is vacuously injective, but if the codomain is non-empty it won't be surjective.

And if the domain is non-empty, then there are no functions with the empty set as codomain.

Arthur
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  • What are we violating in the case where $A$ is nonempty? Why does it only work when $A=B=\emptyset$. – user593295 Dec 13 '20 at 07:55
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    @user593295 "For any $a\in A$, there is a $b\in B$" is part of the definition of a function $A\to B$, and if $A$ is non-empty and $B$ is empty, then that's impossible. – Arthur Dec 13 '20 at 07:59